The Lever when its Weight is taken into Account.— In this case we have to add the moment due to the weight of the lever, to the moment of P or of W according as it acts along with the one force or with the other; i.e., according as the c.g. of the lever is on the same side of the fulcrum as P or W. When the lever is of uniform section and density throughout, then the c.g. of the lever is at its middle point, and consequently the whole weight of the lever may be considered as concentrated and acting at that point. Let AB be a uniform lever, of weight w, acting at its c.g. or middle point C, let a weight, W, be attached to the end B, then the force P, which will have to be applied to the other end A, in order to balance the whole about the fulcrum F, will be found by taking moments about F. Thus, PX AF+w× CF = W × BF P = W x BF-w× CF EXAMPLE I.-A uniform lever, 5 ft. long, of 30 lbs. weight, is placed on a fulcrum 10 in. from one end, and has a weight of 100 lbs. attached to the short end. What force must be applied, and in what direction, in order to produce equilibrium? Also, what is the pressure on the fulcrum, and in what direction does the reaction from the fulcrum act? 1. Referring to the above figure, we find from the question that AB = 5 ft. = 60 in.; BF = 10 in... AF=50 in. and CF = 20 in. W = 100 lbs. and w = 30 lbs. 2. By the principle of moments The Opposing Moments about the Fulcrum are equal. Wx BF-w × CF P= 100 X 10- - 30 x 20 = 8 lbs. Substituting the numerical values— * If the c.g. of the lever was on the opposite side of the fulcrum on the side of W, then P× A=W×BF+w× CF. 3. P acts vertically downwards, since the moment due to the weight of the lever is not sufficient to equalise the moment due to the weight W about the point F. 4. The pressure on the fulcrum is evidently equal to the sum of all the forces, since all the forces act in one direction, or vertically downwards. It is therefore equal to W + w + P = 100 + 30 + 8 = 138 lbs. 5. The reaction from the fulcrum is equal and opposite in direction to this resultant. It therefore acts vertically upwards, and is the equilibrant of the whole of the forces, for a vertical force of 138 lbs. applied to the lever at F would counterpoise or just lift the whole bar with the attached weights P and W. EXAMPLE II. Suppose everything the same as in the previous example but the weight of the lever, which you may consider as now equal to 60 lbs.; what force P would be required, and in what direction would it have to act, in order to produce equilibrium? Also, what would be the resultant or downward pressure at F. 1. You observe at once that the moment of the weight of the lever is greater than the moment of W about the fulcrum. Consequently by the principle of moments P must act against w, or vertically upwards, so as to assist W, in order that the opposing moments about the fulcrum may be equal. Or, 2. The formula therefore becomes wx CF-P x AF=Wx BF wx CFW × BF + P × AF wx CF-W × BF AF P Substituting the numerical values, we have 60 x 20 100 X 10 50 = P=4 lbs. 3. The resultant pressure at F is equal to the algebraical sum of the forces, or = 100 + 60 4 = 156 lbs. W + w Р And acts vertically downwards. The equilibrant would therefore be 156 lbs. acting on the lever at F and vertically upwards. Position of the Fulcrum.-In answering questions which give the magnitude of the forces with which they act, and require only an answer for the position of the fulcrum, the student has simply to employ the general formula for the principle of moments, and then to substitute the known numerical values in order to get the unknown. Or, he may reason out the formula into the following shape, and then interpolate the numerical values. Referring to the last figure, suppose that the distance AF is required: Then, neglecting the weight of the lever, we have by the principle of moments Or, X P × AF = W × BF=W (BA - AF) = W × BA - W × AF. = Now, taking the weight of the lever into account, we have by the principle of moments: PxAF+w (AF-AC)=W (BA-AF) = W x BA - W x AF. Or, == BA PX AF+w × AF + W × AF = W × BA+w× = BA (W+w) EXAMPLE III.-Where should the fulcrum be placed under a uniform lever in order to produce equilibrium, if the lever is 5 ft. long, weighs 30 lbs., and has weights of 100 and 8 lbs. respectively hung at its ends. From the above general equation for equilibrum-viz.: Which proves the data given in Example I. to be correct. LECTURE III.-QUESTIONS. 1. Define what is meant by "the moment of a force," and give an example with a sketch. 2. State "the principle of moments,” and apply it to the case of a simple straight lever. 3. A weight of 10 lbs. on the end of a lever 100 inches from the centre of motion is found to balance a weight of 100 lbs. at a distance of 10 inches. Explain the natural law which governs matter and motion, upon which the above mechanical fact depends. (Answer this by giving the definition of the principle of moments.) 4. Describe an experiment to prove the equality of the moments when the pull is between the weight and the fulcrum and acts in the opposite direction to the weight. 5. In the case of a straight lever, how would you ascertain the pressure on and the reaction from the fulcrum? 6. Three forces, of 12, 10 and 2 lbs., act along parallel lines on a rigid body; show by a sketch how they may be adjusted so as to be in equilibrium? Ans. The force of 12 lbs. must act as the equilibrant to the forces 2 and 10 lbs.-i.e., in a line with their resultant, but in the opposite direction. 7. Two parallel forces of 10 and 12 lbs. act in opposite directions on a rigid body, and at 2 feet apart. Where is the centre of the two forces, and what is their resultant? Ans. 10 feet from the force of 12 lbs., 2 lbs. 8. Define the "centre of gravity" of a body, and show how you would find it experimentally in the case of any irregular body. Give an example. 9. State the rule which applies when two unequal forces balance on opposite sides of the fulcrum of a straight lever, the weight of the lever being neglected. A uniform straight lever, 4 feet long, weighs 10 lbs., the fulcrum is at one end; find what upward force acting at the other end will keep the lever horizontal when a weight of 10 lbs. is hung at a distance of 1 foot from the fulcrum. Find also the pressure on the fulcrum and the direction in which it acts. (S. and A. Exam. 1891.) Ans. 7.5 lbs.; 12.5 lbs. downwards. 10. A uniform bar, 4 feet long and weighing 4 lbs., can turn about a fulcrum at one end, and a weight of 10 lbs. is hung upon the bar at a distance of I foot from the fulcrum. Find the upward force at the free end which will keep the bar horizontal. (S. and A. Exam. 1887.) Ans. 4.5 lbs. 11. A uniform bar of metal 10 inches long weighs 4 lbs., and a weight of 6 lbs. is hung from one end. Find the fulcrum or point upon which the bar will balance. Ans. 2 inches from 6 lbs. 12. Two parallel forces whose magnitudes are 8 and 12 lbs. respectively, act in the same direction on a rigid body at points 10 inches apart. Find the magnitude and line of action of the resultant of the two forces. Ans. 20 lbs. at a point 6 inches from the force of 8 lbs. 13. A uniform lever is 5 feet long, and weighs 10 lbs., the fulcrum being at one end. A weight of 30 lbs. is hung at a distance of 4 feet from the fulcrum; what upward force acting at the middle point of the lever will keep it in a horizontal position? Ans. 58 lbs. LECTURE IV. CONTENTS.-Practical Applications of the Lever-The Steelyard, or Roman Balance-Graduation of the Steelyard-The Lever Safety ValveExample I.—Lever Machine for Testing Tensile Strength of Materials -Straight Levers acted on by Inclined Forces-Bent Levers-The Bell Crank Lever-Bent Lever Balance-Duplex Bent Lever, or Lumberer's Tongs-Turkus, or Pincers-Examples II. and III.-Questions. In this Lecture we shall give a number of examples of the application of the lever. |